Vidinli did not continue after the differences of fourth powers and the sum of the seventh powers. Just out of curiosity, we computed differences of fifth powers, with the following result. \begin{align} S_n^{\,5} - S_{n-1}^{\,\,5} &= {\bigg[ {\frac{n(n+1)}{2}}\bigg]}^5 - {\bigg[ {\frac{(n-1)n}{2}}\bigg]}^5 \\ &= {\frac{n^5}{32}} \Big({(n+1)}^5 - {(n-1)}^5 \Big) \\ &= {\frac{n^5}{32}}(10n^4 +20n^2 + 2) \\ &= {\frac{5}{16}}n^9 +{\frac{5}{8}}n^7 + {\frac{1}{16}}n^5 .\end{align} If we continue the process, we can find a formula for the sum of the ninth powers. Try it! (See Exercise 3.)